Given a finite set of fields $k_1, \ldots, k_n$, is there a (commutative with $1$) ring $R$ with (maximal) ideals $m_i$ such that $R/m_i \cong k_i$?

To prevent things from being too easy, I require two conditions on $R$:

1) $\operatorname{Spec}(R)$ should be connected (otherwise take $R = k_1 \times \ldots \times k_n$)

2) $R$ should be Noetherian (otherwise take $R$ to be a polynomial ring over $\mathbb{Z}$ in sufficiently many variables)

I would be happy with the case $n = 2$ (although I don't currently see how to get the general case from this). However, I do insist that the fields be arbitrary - it is known that any finite collection of countable fields is the set of residue fields of a PID (see this article by Heitmann - first page only).

(I've included the algebraic-geometry tag in hopes for some geometric insight. If however someone feels that this is sufficiently non-geometric, feel free to edit the tags.)

  • 4
    $\begingroup$ The question mathoverflow.net/questions/162030/… and its answers contain partial answers to your question. $\endgroup$ – Olivier Benoist Jul 14 '14 at 20:34
  • 1
    $\begingroup$ @OlivierBenoist: Thank you for linking to that question. However, I don't think that any of the answers there actually resolve this question. The gist seems to be that if condition (1) were upgraded to being a domain, then the answer is negative $\endgroup$ – zcn Jul 14 '14 at 23:06
  • $\begingroup$ It is possible to apply the lemma contained in user46855's answer to this question to the reduced irreducible components of your connected noetherian scheme. Unless I'm mistaken, it proves that the answer to your question is negative if k_1 (say) has cardinality much bigger than k_2. $\endgroup$ – Olivier Benoist Jul 15 '14 at 8:09
  • $\begingroup$ @OlivierBenoist: As far as I can see, this only implies that the two closed points must lie in different irreducible components $\endgroup$ – zcn Jul 15 '14 at 8:11
  • $\begingroup$ Yes but since your scheme is connected, you can join your two points by a chain of connected components and apply the lemma as many times as there are connected components in the chain. $\endgroup$ – Olivier Benoist Jul 15 '14 at 8:18

This answer basically fills in details in Oliver Benoist's comments: If $K$ and $L$ are fields with $|L| > |K|^{\aleph_0}$, then $L$ and $K$ cannot be residue fields of $R$, with $\mathrm{Spec}(R)$ Noetherian and connected.

Proof: Suppose otherwise. Then there is a sequence $K = F_0$, $F_1$, $F_2$, ..., $F_r = L$ of residue fields of $R$, and of irreducible components $\mathrm{Spec}(R_1)$, $\mathrm{Spec}(R_2)$, ..., $\mathrm{Spec}(R_r)$ so that $F_i$ and $F_{i+1}$ are quotients of $R_{i+1}$. By the previous answer, $$|F_r| \leq |F_{r-1}|^{\aleph_0} \leq (|F_{r-2}|^{\aleph_0})^{\aleph_0} \leq \cdot \leq ( \cdots ((|F_0|^{\aleph_0})^{\aleph_0}) \cdots )^{\aleph_0} = |F_0|^{\aleph_0 \times \cdots \times \aleph_0} = |F_0|^{\aleph_0}$$

  • $\begingroup$ Ah, perfect. I must admit it was my own ignorance of cardinal arithmetic that prevented me from seeing this very clean and straightforward argument (i.e. the last two equalities). Many thanks to Olivier Benoist as well. $\endgroup$ – zcn Jul 15 '14 at 16:42
  • $\begingroup$ However, this immediately raises the question: is every Noetherian connected ring a quotient of a Noetherian domain? This answer shows that the type of example given before will not work anymore. Do you (or anyone) see a quick answer, or should this be a separate question? $\endgroup$ – zcn Jul 15 '14 at 16:44
  • $\begingroup$ Not obvious to me, and should definitely be its own question. My guess is yes. $\endgroup$ – David E Speyer Jul 15 '14 at 17:06
  • $\begingroup$ Actually, I take that guess back. There are varieties over $\mathbb{F}_p$ that don't deform to characteristic zero, right? I don't know if there are affine examples, but probably. So take $X$ over $\mathbb{F}_p$ affine, unliftable to $\mathbb{Z}_p$ and with an $\mathbb{F}_p$ point $x$. Glue $X$ to a copy of $\mathrm{Spec} \mathbb{Z}_p$, where the closed point of $\mathrm{Spec} \mathbb{Z}_p$ glues to $x$. How could you embed this in a domain? In any case, it should be its own question. $\endgroup$ – David E Speyer Jul 15 '14 at 17:30
  • $\begingroup$ Section 5 of iu.hio.no/~jank/papers/LiftGraded76.pdf exhibits a complete local noetherian ring over $\mathbb{F}_p$ which does not lift to a complete local noetherian ling over $\mathbb{Z}_p$. I haven't absorbed the example to understand well enough yet to understand what the adjectives complete and local are doing there. If they are removable, then I think we can use this to build a counterexample. $\endgroup$ – David E Speyer Jul 15 '14 at 17:47

Embed every $k_i$ in a huge field $K$, and consider the subring of $k_1[t]\times \ldots \times k_n[t]$ consisting of $n$-uples $(P_1,\ldots ,P_n)$ with $P_1(0)=\ldots =P_n(0)$. Take for $\mathfrak{m}_i$ the ideal of $n$-uples with $P_i(1)=0$.

Edit : As pointed out in the comments, this construction works only in equal characteristic, and doesn't give a noetherian ring in general.

  • $\begingroup$ Note that the $k_i$ may have different characteristics, so such a $K$ need not exist, and then $P_1(0) = \ldots = P_n(0)$ does not make sense anymore $\endgroup$ – zcn Jul 15 '14 at 8:00
  • $\begingroup$ I had great hopes for an example of this type though, and would be happy to see if it can be fixed $\endgroup$ – zcn Jul 15 '14 at 8:07
  • $\begingroup$ Oops, sorry! Indeed this works only if char($k_i$) is constant. $\endgroup$ – abx Jul 15 '14 at 8:13
  • 1
    $\begingroup$ @abx I do not think your ring is going to be noetherian in general. If $n=2$, $k_1=\mathbb{C}$ and $k_2=\mathbb{Q}$, quotienting your ring by the ideal $I=\{(0,Q)\}$, you get the ring $A=\{P\in \mathbb{C}[t]|P(O)\in\mathbb{Q}\}$. But A is not noetherian as its ideal $J=\{P|P(0)=0\}$ is not finitely generated. $\endgroup$ – Olivier Benoist Jul 15 '14 at 8:40
  • $\begingroup$ OK, you are right. $\endgroup$ – abx Jul 15 '14 at 9:49

The answer to your question is basically yes. R. Heitmann proved in his thesis that under mild conditions there exists a PID $R$ with specified residue fields. The conditions will always hold if your collection of residue fields is finite, and each of them are countable.

See: http://www.projecteuclid.org/euclid.dmj/1077310578

  • 2
    $\begingroup$ Nice reference, but your summary of Heitmann's paper is wrong. Hietmann's Theorem A requires the fields to be countable, so it will not "always hold if your collection of residue fields is finite". $\endgroup$ – David E Speyer May 26 '15 at 17:54
  • $\begingroup$ This is also mentioned in the OP. $\endgroup$ – Qiaochu Yuan May 26 '15 at 18:23
  • $\begingroup$ @DavidSpeyer - Thanks for the catch! My fields are almost always countable, so I tend to forget that I'm assuming that. $\endgroup$ – Cory C. May 27 '15 at 14:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.